Question

(i)The lateral or curved surface area of a closed cylindrical petrol storage tank that is $ 4.2m $ in diameter and $ 4.5m $ high.
(ii)How much steel was actually used, if $ \dfrac{1}{{12}} $ of the steel actually used was wasted in making the tank?

Answer 1

Hint: Use the formula of curved surface area of cylinders when height and diameter of the cylinder is given. Just put the value of height and radius in the formula and get the results. For, second part Take out the total surface area of the cylinder for further simplifications.

Complete step-by-step answer:
We are given the height of the cylinder which is, $ h = 4.5m $ and diameter of the base which is, $ d = 4.2m $ .
Take out radius first:
Since we know radius is equal to diameter divided by $ 2 $ .
Radius $ (r) = \dfrac{d}{2} = \dfrac{{4.2}}{2} = 2.1 $
(i)
Put the values of $ r $ and $ h $ in the formula of curved or lateral surface area, which is CSA $ = 2\pi rh $ .
Curved Surface Area of Tank
 $
   = 2\pi rh \\
   = 2 \times \dfrac{{22}}{7} \times 2.1 \times 4.5 \\
   = 2 \times \dfrac{{22}}{7} \times \dfrac{{21}}{{10}} \times \dfrac{{45}}{{10}} \\
   = \dfrac{{22 \times 3 \times 9}}{{10}} \\
   = \dfrac{{594}}{{10}} \\
   = 59.4{m^2} \;
  $
Therefore, the lateral or curved surface area of a closed cylindrical petrol storage tank that is $ 4.2m $ in diameter and $ 4.5m $ high is $ 59.4{m^2} $ .

(ii) Let, the Amount of Steel used in making the tank be x.
And, the $ \dfrac{1}{{12}} $ of the total steel was wasted.
So, the amount of steel wasted $ = \dfrac{1}{{12}} $ of total steel used
 $
   = \dfrac{1}{{12}} \times x \\
   = \dfrac{x}{{12}} \;
  $
Total surface area of the tank $ = $ total of steel used actually
TSA $ = $ (Total steel $ - $ Total steel wasted) ………….(i)
Put the values of $ r $ and $ h $ in the formula Total surface area, which is TSA $ = 2\pi r(r + h) $ .
Total Surface Area of Tank $ = 2\pi r(r + h) $
 $
   = 2 \times \dfrac{{22}}{7} \times 2.1 \times (2.1 + 4.5) \\
   = 2 \times \dfrac{{22}}{7} \times 2.1 \times 6.6 \\
   = 2 \times \dfrac{{22}}{7} \times \dfrac{{21}}{{10}} \times \dfrac{{66}}{{10}} \\
   = \dfrac{{2 \times 22 \times 3 \times 66}}{{10 \times 10}} \\
   = \dfrac{{8712}}{{100}} \\
   = 87.12{m^2} \;
  $
Put the value of TSA in (i):
TSA $ = $ (Total steel $ - $ Total steel wasted)
 $
  87.12 = x - \dfrac{x}{{12}} \\
  87.12 = \dfrac{{12x - x}}{{12}} \\
  87.12 = \dfrac{{11x}}{{12}} \\
  87.12 \times 12 = 11x \\
  1045.44 = 11x \\
  \dfrac{{1045.44}}{{11}} = x \\
   = > x = 95.04{m^2} \;
  $
Therefore, the amount of steel was actually used, if $ \dfrac{1}{{12}} $ of the steel actually used was wasted in making the tank $ 95.04{m^2} $ .
So, the correct answer is “ $ 95.04{m^2} $ ”.

Note: Always remember the formulas for Curved surface or Total surface area of cylinder.
Do not use diameter instead of radius in the formula, check whether it's given radius or diameter in the question.
Don’t use only $ \dfrac{1}{{12}} $ instead of \[\dfrac{x}{{12}}\] otherwise it would result in error.